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Question

If |z|1,|w|1, then

A
|zw|2=(|z||w|)2+(argzargw)2
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B
|zw|2(|z||w|)2+(argzargw)2
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C
|zw|2(|z||w|)2+(argzargw)2
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D
None of the above
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Solution

The correct option is B |zw|2(|z||w|)2+(argzargw)2
We know that, |zw|2=|z|2+|w|22Re(z ¯¯¯¯w )
=|z|2+|w|22|z||w|cos(αβ),
(where α=argz, β=argw)
So, |zw|2=(|z||w|)2+2|z||w|(1cos(αβ))
(|z||w|)2+4sin2(αβ2)2 (|z|1,|w|1)
(|z||w|)2+4(αβ2)2 (sinθ<θ for θ (0,π2))
(|z||w|)2+(αβ)2
(|z||w|)2+(argzargw)2

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